NOT for inverting bits in Embedded C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the NOT operator (~) do in embedded C?

The NOT operator (~) flips every bit of a number: 0 becomes 1, and 1 becomes 0. It's used to invert bits.

beginner

How do you invert all bits of an 8-bit number 0b10101010 using NOT (~)?

Applying ~0b10101010 results in 0b01010101, flipping each bit from 1 to 0 or 0 to 1.

intermediate

Why is NOT (~) useful in embedded programming?

It helps quickly toggle bits, create masks, or invert flags without loops, saving time and code space.

intermediate

What happens if you apply NOT (~) to an unsigned int with value 0?

All bits become 1, so the result is the maximum unsigned int value (e.g., 0xFFFFFFFF for 32-bit).

advanced

Can NOT (~) be used to invert only some bits of a number?

No, NOT (~) inverts all bits. To invert specific bits, combine NOT with masks and bitwise AND/OR.

What does the ~ operator do in embedded C?

A Shifts bits to the right

B Shifts bits to the left

C Sets all bits to zero

D Inverts all bits of a number

The ~ operator flips every bit: 0 becomes 1, and 1 becomes 0.

If x = 0b00001111, what is ~x in 8-bit?

A 0b11110000

B 0b00001111

C 0b11111111

D 0b00000000

NOT flips bits, so 00001111 becomes 11110000.

Can ~ operator invert only selected bits directly?

A Yes, it can invert any chosen bits

B No, it inverts all bits

C Only if combined with addition

D Only on signed integers

~ always inverts all bits of the operand.

What is the result of ~0 for a 16-bit unsigned integer?

A 0x0000

B 0x7FFF

C 0xFFFF

D 0x8000

~0 flips all zeros to ones, so all bits become 1 (0xFFFF).

Why might embedded programmers use ~ operator?

A To invert bits quickly without loops

B To multiply numbers

C To divide numbers

D To convert numbers to strings

~ operator is a fast way to flip bits directly.

Explain how the NOT (~) operator works to invert bits in embedded C.

Think about how each bit changes when you apply ~.

Describe a situation in embedded programming where using the NOT (~) operator is helpful.

Consider how flipping bits can simplify code.

Practice

1.

What does the ~ operator do in embedded C?

easy

A. It flips every bit of the operand (bitwise NOT).

B. It shifts bits to the left by one position.

C. It adds one to the operand.

D. It compares two values for equality.

Solution

1. Step 1: Understand the ~ operator

   The ~ operator is the bitwise NOT operator in C, which flips every bit of its operand.

2. Step 2: Compare with other options

   Other options describe different operations: shifting, addition, and comparison, which are not what ~ does.

3. Final Answer:

   It flips every bit of the operand (bitwise NOT). -> Option A

4. Quick Check:

   ~ operator = bitwise NOT [OK]

Hint: Remember ~ flips bits, like turning 0s to 1s and vice versa [OK]

Common Mistakes:

• Confusing ~ with shift operators
• Thinking ~ adds or subtracts
• Mixing ~ with logical NOT (!)

2.

Which of the following is the correct syntax to invert bits of an unsigned integer x in embedded C?

unsigned int x = 5;
unsigned int y = ?;

easy

A. y = x++;

B. y = !x;

C. y = ~x;

D. y = x & ~x;

Solution

1. Step 1: Identify the bit inversion syntax

   To invert bits, use the bitwise NOT operator ~ directly on the variable.

2. Step 2: Check other options for correctness

   y = !x; uses logical NOT, which returns 0 or 1, not bit inversion. y = x++; increments x, not invert bits. y = x & ~x; ANDs x with its inverse, resulting in zero.

3. Final Answer:

   y = ~x; -> Option C

4. Quick Check:

   Bit inversion syntax = y = ~x; [OK]

Hint: Use ~ directly on variable to invert bits [OK]

Common Mistakes:

• Using ! instead of ~
• Using increment operators by mistake
• Combining & with ~ incorrectly

3.

What is the output of the following code snippet?

unsigned char x = 0b00001111;
unsigned char y = ~x;
printf("%u", y);

medium

A. 240

B. 15

C. 0

D. 255

Solution

1. Step 1: Understand the initial value of x

   x is 0b00001111, which is decimal 15.

2. Step 2: Apply bitwise NOT to x

   ~x flips bits: 00001111 becomes 11110000 in 8 bits, which is decimal 240.

3. Final Answer:

   240 -> Option A

4. Quick Check:

   ~15 (8-bit) = 240 [OK]

Hint: Invert bits of 00001111 to get 11110000 = 240 decimal [OK]

Common Mistakes:

• Confusing logical NOT (!) with bitwise NOT (~)
• Forgetting unsigned type size
• Assuming output is 15 or 255

4.

Find the error in the following code that attempts to invert bits of num and print the result:

int num = 5;
int inverted = ~num;
printf("%u", inverted);

medium

A. The variable inverted should be unsigned char.

B. The ~ operator cannot be used on int types.

C. Missing semicolon after variable declaration.

D. Using %u to print a signed int causes incorrect output.

Solution

1. Step 1: Check the type and format specifier

   Variable inverted is signed int, but printed with %u (unsigned). This mismatch can cause wrong output.

2. Step 2: Verify other options

   The ~ operator works on int types. Semicolons are present. Changing to unsigned char is not necessary but can clarify output.

3. Final Answer:

   Using %u to print a signed int causes incorrect output. -> Option D

4. Quick Check:

   Format specifier must match variable type [OK]

Hint: Match printf format (%d for int, %u for unsigned) [OK]

Common Mistakes:

• Using wrong printf format specifier
• Thinking ~ can't be used on int
• Ignoring signed vs unsigned mismatch

5.

You want to invert bits of a 16-bit unsigned integer val but only for the lower 8 bits, leaving the upper 8 bits unchanged. Which code snippet correctly does this?

unsigned short val = 0xABCD;
unsigned short result = ?;

hard

A. result = ~val & 0xFFFF;

B. result = (val & 0xFF00) | (~val & 0x00FF);

C. result = ~val | 0xFF00;

D. result = val ^ 0xFFFF;

Solution

1. Step 1: Understand the requirement

   Invert only the lower 8 bits of val, keep upper 8 bits as is.

2. Step 2: Analyze each option

   result = (val & 0xFF00) | (~val & 0x00FF); keeps upper 8 bits (val & 0xFF00) and inverts lower 8 bits (~val & 0x00FF), then combines them. result = ~val & 0xFFFF; inverts all 16 bits. result = ~val | 0xFF00; incorrectly sets upper bits to 1. result = val ^ 0xFFFF; flips all bits (XOR with 0xFFFF).

3. Final Answer:

   result = (val & 0xFF00) | (~val & 0x00FF); -> Option B

4. Quick Check:

   Mask upper bits, invert lower bits, combine [OK]

Hint: Mask upper bits, invert lower bits, then combine [OK]

Common Mistakes:

• Inverting all bits instead of just lower 8
• Using OR instead of AND for masking
• Forgetting to combine masked parts